126 .instructionset
We’ve now reversed all the VM instructions, and have a full understanding about how it works. Here is the VM’s instruction set:
| Instruction | 1st arg | 2nd arg | What does it do? |
|---|---|---|---|
| “A” | “M” | arg2 | *sym.current_memory_ptr += arg2 |
| “P” | arg2 | sym.current_memory_ptr += arg2 | |
| “C” | arg2 | sym.written_by_instr_C += arg2 | |
| “S” | “M” | arg2 | *sym.current_memory_ptr -= arg2 |
| “P” | arg2 | sym.current_memory_ptr -= arg2 | |
| “C” | arg2 | sym.written_by_instr_C -= arg2 | |
| “I” | arg1 | n/a | instr_A(arg1, 1) |
| “D” | arg1 | n/a | instr_S(arg1, 1) |
| “P” | arg1 | n/a | *sym.current_memory_ptr = arg1; instr_I(“P”) |
| “X” | arg1 | n/a | *sym.current_memory_ptr ^= arg1 |
| “J” | arg1 | n/a | arg1_and_0x3f = arg1 & 0x3f; if (arg1 & 0x40 != 0) arg1_and_0x3f *= -1 if (arg1 >= 0) return arg1_and_0x3f; else if (*sym.written_by_instr_C != 0) { if (arg1_and_0x3f < 0) ++*sym.good_if_ne_zero; return arg1_and_0x3f; } else return 2; |
| “C” | arg1 | n/a | *sym.written_by_instr_C = arg1 |
| “R” | arg1 | n/a | return(arg1) |